A buggy is pulling a roller-skater, in a straight line along a horizontal road, by means of a connecting rope as shown in the diagram - AQA - A-Level Maths Mechanics - Question 17 - 2018 - Paper 2

The tension T in the rope can again be determined using the roller-skater's forces. For the roller-skater, the equation of motion is:

$T - R = m \cdot a$

Inserting the known values:

$T - 63.6 \, N = 72 \, kg \cdot 0.2 \, m/s^2$

This simplifies to:

$T - 63.6 \, N = 14.4 \, N$

Thus,

$T = 63.6 \, N + 14.4 \, N = 78 \, N$

A necessary assumption made in this calculation is that the rope has no mass or that it is horizontal and inextensible.

To determine if the roller-skater will stop before the buggy, we first need to calculate the deceleration experienced by the roller-skater after releasing the rope:

The resistance force acting on her remains R, which is 63.6 N.

Using the formula for acceleration:

$F = m \cdot a \, \Rightarrow \quad a = \frac{F}{m} = \frac{-63.6 \, N}{72 \, kg} = -0.883 \, m/s^2$

We now apply the equation of motion:

$v^2 = u^2 + 2as$

Where: v = final velocity = 0 (when she stops) u = initial velocity = 6 m/s a = -0.883 m/s²

Setting s as the distance she travels while stopping:

$0 = (6)^2 + 2(-0.883)s$

Solving for s gives:

$s = \frac{36}{2 \times 0.883} = 20.4 \, m$

Therefore, since 20.4 m > 20 m, she will stop after traveling 20.4 m, which means she will not stop before reaching the stationary buggy.

The driver noticed a change in motion of the buggy because, upon the roller-skater releasing the rope, the resistance force acting against the roller-skater increased due to her continued forward motion. As a result, the buggy's acceleration decreased significantly, as the force exerted on the buggy was no longer complemented by the force exerted by the roller-skater. This observed change would appear as an immediate deceleration of the buggy as it no longer experienced the same dynamics originally in play when the roller-skater was still attached.